course Mth 173 {ރP~ݣJ۽assignment #006
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14:28:30 `qNote that there are 7 questions in this assignment. `q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?
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RESPONSE --> Asuming that the rate of depth change is still around -4cm/sec, which it should be, because in the one second the rate shouldn't change by an extremely large amount. So if the change in time is one second and the depth at Ti was 80cm then the depth at Ti + 1= 21=T should be something around 76cm. confidence assessment: 2
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14:29:19 At a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm.
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RESPONSE --> okay, I might have read into that question a little bit. self critique assessment: 3
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14:36:00 `q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant?
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RESPONSE --> From the original point of 80cm at 20 sec. we have progressed 10sec. so the water has dropped 40cm. 80-40=40 so at 30sec. we would have 40cm of water left. This estimation is less accurate because there is a constantly changing rate of depth change, so we may be off by a few cm. over this much time. confidence assessment: 2
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14:36:31 At - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm.
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RESPONSE --> okay self critique assessment: 3
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14:44:28 `q003. If you know that the depth in the preceding example is changing at the rate of -3 cm/s at the t = 30 sec instant, how will this change your estimate for the depth at t = 30 seconds--i.e., will your estimate be the same as before, will you estimate a greater change in depth or a lesser change in depth?
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RESPONSE --> knowing that the rate of change changed during the 10sec. period it will throw the prevous answer because of the different rates of change, with out a model you could find the average of the two rates and use that in our calculations. -3.5cm/sec. average rate of change, and for 10s. so the depth changing by -35cm subtracted from the original 80 cm. is 45cm at time 30 with a changing rate. confidence assessment: 2
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14:45:40 Since the rate of depth change has changed from -4 cm / s at t = 20 s to -3 cm / s at t = 30 s, we conclude that the depth probably wouldn't change as much has before.
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RESPONSE --> The depth would not have chnaged as much as before, originally it changed 40cm but then with the different change it only changed 35cm. self critique assessment: 2
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14:47:14 `q004. What is your specific estimate of the depth at t = 30 seconds?
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RESPONSE --> my guess using the average rate of depth change from -4 to -3, is when the time is 30 sec. then the depth is 45 cm. confidence assessment: 2
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14:47:50 Knowing that at t = 20 sec the rate is -4 cm/s, and at t = 30 sec the rate is -3 cm/s, we could reasonably conjecture that the approximate average rate of change between these to clock times must be about -3.5 cm/s. Over the 10-second interval between t = 20 s and t = 30 s, this would result in a depth change of -3.5 cm/s * 10 sec = -35 cm, and a t = 30 s depth of 80 cm - 35 cm = 45 cm.
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RESPONSE --> The longer version of the work for this problem was included in the last problem. self critique assessment: 3
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14:55:18 `q005. If we have a uniform cylinder with a uniformly sized hole from which water is leaking, so that the quadratic model is very nearly a precise model of what actually happens, then the prediction that the depth will change and average rate of -3.5 cm/sec is accurate. This is because the rate at which the water depth changes will in this case be a linear function of clock time, and the average value of a linear function between two clock times must be equal to the average of its values at those to clock times. If y is the function that tells us the depth of the water as a function of clock time, then we let y ' stand for the function that tells us the rate at which depth changes as a function of clock time. If the rate at which depth changes is accurately modeled by the linear function y ' = .1 t - 6, with t in sec and y in cm/s, verify that the rates at t = 20 sec and t = 30 sec are indeed -4 cm/s and -3 cm/s.
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RESPONSE --> By plugging in the times to the function y' = .1t-6 we get the data set of. {(20, -4.0), (21, -3.9), (30, -3.0)} which correctly correspond with the given rates. confidence assessment: 3
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14:55:31 At t = 20 sec, we evaluate y ' to obtain y ' = .1 ( 20 sec) - 6 = 2 - 6 = -4, representing -4 cm/s. At t = 30 sec, we evaluate y' to obtain y' = .1 ( 30 sec) - 6 = 3 - 6 = -3, representing -3 cm/s.
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RESPONSE --> okay self critique assessment: 3
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14:57:49 `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero?
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RESPONSE --> to find when the rate first equals zero. We wokr the equation: 0=.1t-6 6=.1t 60=t So at t=60s. is the first time that the rate of change equals 0. confidence assessment: 3
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14:57:57 The rate of depth change is first equal to zero when y ' = .1 t - 6 = 0. This equation is easily solved to see that t = 60 sec.
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RESPONSE --> okay self critique assessment: 3
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15:02:15 `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing?
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RESPONSE --> The rate of change at t=20 is -4cm/s while the rate of change at t=60 is 0cm/s so the average rate of depth change during this intrval is -2cm/s. since there is 40s. of time passing 40s * -2cm/s= -80cm so there is a depth change of 80 cm from t=20 untill the rate reaches 0. confidence assessment: 3
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15:03:05 The rate of depth change at t = 20 sec is - 4 cm/s; at t = 60 sec the rate is 0 cm/s. The average rate at which depth changes during this 40-second interval is therefore the average of -4 cm/s and 0 cm/s, or -2 cm/s. At an average rate of -2 cm/s for 40 s, the depth change will be -80 cm. Starting at 80 cm when t = 20 sec, we see that the depth at t = 60 is therefore 80 cm - 80 cm = 0.
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RESPONSE --> okay self critique assessment: 3
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